ENGINEERING  LIBRARY 


A  GRAPHICAL  TREATMENT 


OF  THE 


INDUCTION  MOTOR 

""0 


Translated  from  the  Second  Edition 
G.  H.  ROWE  and  R.  E.  HELLMUND 


NEW  YORK 

McGraw  Publishing  Company 


1906 


&NGINEERIMG  LIBRARY 

Copyrighted,  1906, 

oy  tne 

McGRAW  PUBLISHING  COMPANY 
New  York 


CONTENTS. 

PAGE 

Introduction 1 

General  Theory 1 

Diagrams  of  the  Induction  Motor. 5 

Current  Diagram 6 

Field  Diagram 7 

Circle  Diagram 8 

Determination  of  Input,  etc.,  from  the  Diagram 10 

Friction  and  Iron  Losses 10 

Copper  Losses 10 

Electrical  Input 13 

Torque 13 

Output 14 

Slip 14 

Electrical  Efficiency 15 

Practical  Application  of  the  Diagram 17 

Note  on  Correction  to  be  Applied 22 

Examples 23 

2  H.P.  Motor 24 

7  H.P.  Motor 29 

12  H.P.  Motor. 33 

Induction  Motor  as  Generator 34 

Single  Phase  Motor. 36 


838991 


A  GRAPHICAL  TREATMENT 
OF  THE  INDUCTION  MOTOR. 


The  object  of  the  method  described  in  the  following 
pages  is  the  experimental  determination  of  the  charac- 
teristic properties  of  induction  motors.  It  consists  es- 
sentially in  the  practical  application  of  the  circle  dia- 
gram, first  described  by  the  writer  in  1894.*  It  is 
based  on  two  simple  and  quickly  performed  experimen- 
tal tests  on  the  finished  motor. 

The  method  shows  at  a  glance  the  main  properties 
of  a  motor  and  its  commercial  excellence.  The  writer 
has  used  the  circle  diagram  several  years  in  the  testing 
room  for  comparing  calculated  values  with  results  ob- 
tained from  tests,  and  it  has  well  served  its  purpose. 

Before  describing  the  method  and  its  applications, 
perhaps  I  may  be  allowed  to  present  briefly  the  theory 
of  the  induction  motor,  and  the  derivation  cf  the 
diagram. 

GENERAL  THEORY  OF  THE  INDUCTION  MOTOR. 

The  induction  motor  is  in  principle  a  transformer. 
The  exciting  member  A  (Fig.  1)  represents  the  inducing 
or  primary  circuit ;  the  short-circuited  member  B  repre- 

*Elektrotechnische  Zeitschrift,  Oct.  11,  1894,  p.  561. 

1 


2  THE  INDUCTION  MOTOR. 

sents  the  induced  or  secondary  circuit.  The  alter- 
nating fields  produced  by  the  current  in  the  exciting 
coils  combine  in  the  well  known  manner  to  form  a 
rotating  field.  The  motor  is,  therefore,  a  transformer 
with  a  rotating  field,  and  the  load  of  the  system  is  deter- 
mined at  any  time  by  the  difference  between  the  con- 
stant speed  of  rotation  of  the  field  produced  in  the 
stationary  member  .4,  and  that  of  the  rotating  short- 
circuited  member  B.  The  turning  of  the  rotor  is  due 


FIG.  1. 

to  the  torque  existing  between  the  rotating  field  and 
the  currents  produced  in  the  short-circuited  member 
by  rotation  in  the  field.  These  currents,  and  therefore 
also  the  torque,  vary  directly  with  the  product  of  the 
slip  and  the  field  interlinked  with  the  secondary  con- 
ductors. 

The  slip  is  defined  as  the  difference  between  the 
speed  of  rotation  of  the  exciting  field,  and  that  of  the 
short-circuited  secondary.  The  variation  of  the  slip  in 


THE  INDUCTION  MOTOR.  3 

the  induction  motor  has  the  same  significance  as  the 
variation  of  load  of  the  ordinary  transformer,  due  to 
a  change  of  its  external  resistance. 

With  constant  impressed  electromotive  force,  the 
field  produced  by  the  current  in  the  primary  winding 
is  constant  for  all  loads,  as  in  the  transformer.  This, 
however,  is  not  true  in  the  short-circuited  member, 
which  is  the  real  work  transmitting  element.  Herein 
lies  the  essential  difference  between  the  induction  motor 
and  the  transformer  without  leakage,  in  which,  as  is 
well  known,  the  total  primary  field  passes  through  the 
secondary.  Since  there  must,  of  necessity,  be  an  air- 
gap  between  the  short  circuited  windings  and  the  pri- 
mary windings,  not  all  of  the  magnetic  lines  <£  induced 
in  the  primary  member  pass  through  into  the  secondary. 
But  a  part  <f>s  passes  directly  through  the  space  be- 
tween the  two  windings  back  into  the  primary  mem- 
ber, and  only  the  difference  <£A  =  <fi  —  <ps  passes  into 
the  short-circuited  member.  This  field  </)A  produces,  in 
the  latter,  currents  which  lag,  according  to  the  law  of 
induction,  a  quarter  period  behind  the  field,  i.e.,  they 
are  in  quadrature  with  the  inducing  field,  and  there- 
fore the  torque  produced  is  proportional  to  the  pro- 
duct of  induced  currents  and  primary  field. 

The  lines  <ps  are  called  leakage  lines  or  leakage  field, 
and  since  they  are  caused  by  the  primary  current  7t 

*In  fact  there  is  not  only  a  primary  leakage,  but  also  a 
secondary  leakage  that  is  a  flux  which  is  interlinked  with  the 
secondary  conductors  only.  It  is  however  admirable,  and 
simplifies  further  considerations,  to  assume  that  the  primary 
and  secondary  leakage  fluxes  are  combined  to  form  one  single 
leakage  flux — i.  e.  the  primary  leakage  flux  0S.  The  inexact- 
ness introduced  thereby  is  of  no  practical  importance  in  the 
results  obtained  from  the  following  derivation. — ROWE  HELL- 
MUND. 


4  THE  INDUCTION  MOTOR. 

only,  they  must  be  proportional  to  it.     If  we  designate 
the  reluctance  of  the  leakage  path  by  ps,  then  we  have 


The  main  field,  as  in  the  transformer,  is  of  constant 
amplitude  =  <£.  Therefore,  the  armature  field  ^>A  is 
given  by  the  difference  between  the  main  field  and 
the  leakage  field  <£A  =  <£  -  </>*  (Fig.  2),  and  is  pro- 


FIG.  2. 

duced  by  the  difference  between  the  ampere  turns  of 
the  primary  and  secondary  circuits.  If  we  call  the 
resulting  magnetizing  current,  I,  and  the  reluctance 
of  the  path  of  the  secondary  field  p,  then 

/ 
<£Ax- 

This  magnetizing  current  is  thus  not  constant,  as 
in  the  transformer,  but  decreases  with  increasing  load 
since  with  increasing  load  the  current  7t  and  therefore 
c6c  increases.  </>A  will  therefore  decrease. 


THE  INDUCTION  MOTOR.  j 

If  no  current  is  produced  in  the  secondary  mem- 
ber, the  primary  winding  carries  only  the  magnetizing 
current,  and  therefore  the  primarv  current  equals  the 
magnetizing  current,  or 

/•--/ 

This  occurs  when  the  motor  is  running  unloaded,  and 
the  slip  is  zero. 

As  the  load  increases,  we  have  the  following  phe- 
nomena: The  slip  increases,  and  with  it  the  electro- 
motive force  induced  in  the  secondary  member,  and 
the  current  72  resulting  therefrom  increases.  This 
secondary  current  72  exerts  a  demagnetizing  effect  on 
the  field,  and  therefore  causes,  as  in  the  transformer, 
a  corresponding  increase  of  the  current  in  the  primary 
member.  The  demagnetizing  influence  of  the  current 
7  2  acts  only  on  the  armature  field.  The  strength  of  the 
leakage  field  increases,  however,  in  direct  ratio  with 
the  increase  of  the  primary  current.  Therefore,  the 
increase  of  the  load,  and  the  increase  of  the  secondary 
current,  have  the  following  consequences: 

(1)  The  secondary  field  is  decreased  by  the  demag- 
netizing action  of  72. 

(2)  Indirectly,  the  leakage  field  is  increased,  due  to 
an  increase  of  7X. 

Thus,  the  leakage  field  is  directly  proportional  to 
Iv  The  armature  field  equals  the  difference  between 
the  main  and  leakage  fields,  and  therefore  the  sum  of 
the  armature  field  and  leakage  field  must  be  equal 
to  the  main  field  and  constant. 

DIAGRAM     OF    THE     INDUCTION    MOTOR. 

As  in  the  well  known  transformer  diagram,  the  pri- 
mary current,  and  the  secondary  current  in  the  short- 
circuited  secondary,  can  be  represented  by  two  sides 


6  THE  INDUCTION  MOTOR. 

of  a  triangle  (Fig.  3),  of  which  the  resulting  third  side 
is  the  magnetizing  current  7. 

This  magnetizing  current  produces  the  secondary 
field,  which  interlinks  with  both  the  exciting  and  short- 
circuited  windings.  The  secondary  field  produces  the 
currents  in  the  short-circuited  secondary;  it  being  in 
fact  the  only  field  which  passes  into  the  secondary. 
The  secondary  currents,  therefore,  lag  exactly  a  quarter 
period  behind  the  secondary  field.  As  shown  before, 
the  main  difference  between  the  transformer  without 
leakage  and  the  induction  motor,  lies  in  the  fact  that 


Impressed  E.M.F. 
FIG.   3. 

the  magnitude  of  the  magnetizing  current  changes,  and 
that  the  current  triangle  changes  its  position. 

We  know  (Fig.  3)  that  the  constant  main  field,  which 
induces  a  counter  electromotive  force  corresponding  to 
the  impressed  electromotive  force  is  at  right  angles  to 
the  impressed  electromotive  force  in  a  circuit  without 
resistance,  and  without  iron  losses.  Under  these  con- 
ditions, in  a  transformer  without  leakage,  the  magnet- 
izing current  is  therefore  at  right  angles  to  the  im- 
pressed electromotive  force,  and  is  constant. 

Further,  the  current  triangle  always  has  the  same 
relative  position  to  the  electromotiv?  force. 


THE  INDUCTION  MOTOR  7 

In  the  induction  motor,  nowever,  the  main  field  must 
be  resolved  into  two  parts.  One  part,  the  leakage  field, 
is  in  phase  with  the  primary  current,  and  proportional 
to  it.  The  other  part,  the  secondary  field,  must  there- 
fore be  variable,  and  cannot  be  in  phase  with  the 
main  field.  The  magnetizing  current  which  produces 
this  secondary  field,  must  therefore  vary  in  amplitude 
and  direction.  It  is  possible  to  follow  these  changes 
by  the  graphical  combination  of  the  fields  (Fig.  4). 

The   diagram  proposed  by  the   writer  is  based  on 


Impressed  E.M.F. 


FIG.   4. 

these  assumptions,  and  is  represented  in  Fig.  5. 

The  leakage  field  is  directly  proportional  to  the  pri- 
mary current,  and  in  phase  with  it,  or 


<f>s  oc^-  =  AC' 

Ps 

Thus,  we  can  represent  the  leakage  field  according  to 
a  definite  scale  by  the  vector  AC'  =  Ir  The  main 
field  may  be  represented  by  a  line  A  D,  constant  in 


8 


THE  INDUCTION  MOTOR. 


magnitude  and  in  quadrature  with  the  electromotive 
force. 


The  third  side  C'  D,  must  therefore  give  the  second- 
ary field,  or 

<£A  =—  =  C'L> 
and  C'  D  must  be  in  quadrature  with  c'  C'  =  /2,  and 

D 


Impressed  E.M.F. 


FIG.  5. 


parallel  with  A  c'  =  I.  Further,  there  must  be  a  con- 
stant relation  between  C'  D  and  A  c'  according  to  an 
equation  previously  given,  or 


C'D_<f>       I 

~A T      ~          — T~  &•    — 

Ac'         I       o 


In    order   to    satisfy   this   equation   when   the   load 
changes,  the  points  c'  and  C'  must  move  on  the  two 


THE  INDUCTION  MOTOR.  9 

half  circles  represented  in  Fig.   5.     The  angles  A  c'  C 
and    C  C'  D   are    always    right   angles,    and    the    ratio 

C'D  . 

- — -  is  constant. 

We  see  from  the  figure  that  the  current  cannot  in- 
crease indefinitely,  even  if  we  assume  the  resistance  to 
be  zero.  It  reaches  a  maximum  when  C'  coincides 

with  D,  or 

I,  =  AD. 

The  leakage  field,  which  is  then  represented  by  the 
line  A  D,  equals  in  intensity  the  main  field,  and  the 
remaining  part,  the  secondary  field,  becomes  zero. 

In  this  case 

/!       AD 

(h    =   (h    OC  — -   OC  

PS  Ps 

Let  us  study  now  the  condition  for  minimum  current. 
The  minimum  current  flows  when  the  motor  is  run- 
ning without  load.  In  this  case  72  =  0,  and  therefore 
/j  =  the  magnetizing  current,  or 

7t  =  /  =  A  C,  and 

I        I,     AC 

d>  oc  —  oc  —  oc 

P       P  P 

or,  since  <£  remains  constant,  we  have  the  relation: 

Maximum  current        A  D        ps 
Minimum  current        AC          p 

reluctance  of  leakage  path 


reluctance  of  path  of  secondary  field 

The  above  considerations  may  be  briefly  stated  as 
follows:  In  induction  motors  the  relation  between  the 


JO  THE  INDUCTION  MOTOR. 

electromotive  force  and  the  current  can  be  represented 
by  a  vector  diagram.  In  this  diagram,  the  vector 
representing  the  current,  which,  as  we  know,  varies 
with  the  load,  is  determined  by  the  fact  that  its  end 
point  moves  in  a  circle;  the  position  of  which  is  given 
by  the  ratio 

A  D       jOs  _  reluctance  of  leakage  path 

AC         p        reluctance  of  path  of  secondary  field 

DETERMINATION    OF    INPUT,    OUTPUT,    TORQUE    AND    SLIP. 

So  far,  we  have  investigated  current  magnitudes  and 
phase  differences  only.  To  utilize  the  diagram,  prac- 
tically, and  to  determine  from  it  the  operation  and 
commercial  excellence  of  a  motor,  we  must  introduce 
the  friction,  iron  losses,  and  especially  the  electrical 
losses. 

The  friction  and  iron  losses  can  be  considered  con- 
stant, since  the  speed  of  the  motor  and  the  field  be- 
tween no  load  and  full  load  are  practically  constant. 
The  iron  losses  of  the  short-circuited  secondary  can  be 
neglected  on  account  of  the  low  frequency  of  its  field. 
We  shall  see  later  that  the  exciting  field  does  not 
remain  quite  constant,  but  is  somewhat  reduced  by 
the  ohmic  drop  in  the  primary  circuit.  On  the  other 
hand,  the  iron  losses  in  the  secondary  increase  slightly 
with  the  load,  due  to  increasing  slip.  These  variations 
will  almost  neutralize  each  other,  so  that  the  total 
:'ron  losses  are  approximately  constant. 

It  is  quite  different  with  the   copper  losses.     As  is 
well   known,   these   are   proportional  to   the   square   of 
the   current.     For  this   reason,   it   is   very   difficult   to 
represent  them  directly  in  the  diagram.     We  shall  see 
however,  that  their  influence  appears  in  another  form, 


THE  INDUCTION  MOTOR.  11 

viz.,  by  weakening  the  field,  whereby  they  may  be 
very  readily  considered.  The  ohmic  drop  of  the  pri- 
mary winding — I^xR^  (7t  =  current,  and  R^  =  ohmic 
resistance  of  primary  circuit)  opposes  the  primary  im- 
pressed electromotive  force,  and  consequently  the  field 
does  not  need  to  induce  a  counter  electromotive  force 
equal  to  the  impressed,  but  one  equal  to  the  difference 
between  the  impressed  electromotive  force  and  the 
drop.  It  follows  that  the  main  field,  inducing  the 
counter-electromotive  force,  and,  therefore,  also  the 
secondary  field,  will  be  decreased  by  an  amount  corre- 
sponding to  the  ohmic  drop.  The  diagram  was,  how- 
ever, constructed  on  the  assumption  of  a  constant 
main  field.  Yet  the  omission  of  the  loss  of  potential, 
due  to  resistance,  does  not  change  the  correctness  of 
the  diagram,  since,  instead  of  the  drop,  we  can  intro- 
duce an  equivalent  field,  which  may  finally  be  sub- 
tracted from  the  resulting  seconda^  field. 

Since  an  electromotive  force  always  corresponds  to 
a  field  at  right  angles  to  itself,  and  is  proportional  to 
it,  we  may  take  the  ohmic  drop  into  consideration  by 
a  reduction  of  the  field  proportional  to  and  at  right 
angles  to  the  drop,  that  is  at  right  angles  to  the  direc- 
tion of  the  current.  Thus,  we  arrive  at  correct  results 
if  we  consider  the  main  field  constant  in  our  diagram, 
and  finally  decrease  the  secondary  field  by  an  amount 
corresponding  to  the  drop  in  potential. 

In  the  diagram,  the  primary  current  A  C'  may  be 
resolved  into  two  components,  A  C  and  C  C't  of  which 
the  no  load  component  A  C  causes  a  constant  drop. 
The  other  component  C  C'  is  variable,  and  causes  a 
drop  in  potential  which  appears  in  the  form  of  a  dimi- 
nution of  the  field  <pa  proportional  to  and  in  quadrature 
with  this  component.  In  other  words,  since  part  of 


12 


THE  INDUCTION  MOTOR. 


tha  impressed  electromotive  force  is  used  in  overcoming 
the  ohmic  resistance,  the  secondary  field  needs  only 
to  be  assumed  smaller  by  an  amount  corresponding  to 
the  ohmic  drop.  Let  us  call  this  part  to  be  subtracted 


FIG.  6. 

C'  E'\  then  Cf  E'  is  proportional  to  CC'\  that  is,  the 
point  Er  must  move  on  the  circumference  of  the  second 
circle  shown  in  Fig.  6.  This  circle,  therefore,  indicates 
that  the  secondary  field  is  not  represented  as  before 


THE  INDUCTION  MOTOR.  13 

by  DC',  but  by  the  line  D  Ef  The  main  field  de- 
creases, of  course,  by  the  same  amount.  However, 
since  we  do  not  use  the  main  field  in  further  investiga- 
tion, it  is  unnecessary  to  introduce  the  change  in  the 
diagram. 

Similarly,  the  loss  in  the  short-circuited  secondary 
appears  in  the  diagram  as  the  line  E'  F'.  Introducing 
in  this  way  the  ohmic  drop  of  both  windings  by  a  corre- 
sponding reduction  of  the  secondary  field,  we  are  in  a 
position  to  derive  the  various  quantities,  such  as 
torque,  output,  etc. 

(a)  The  electrical  input  =  3  e  i  cos  6  =  v7  3  E  i  cos  0, 
can  be  found,  since  the  impressed  electromotive  force 
is  constant,  by  scaling  the  line  c'  C',  that  is,  the  energy 
component  of  the  primary  current  A  C', 

.  '  .  Input  oc  c'  C' 

(b)  The  torque  is  determined  by  the  product  of  the 
secondary  field  and  the  secondary  current,  and  is  pro- 
portional to  E'  D  x  C'  C,  i.e.,  to  the  area  of  the  triangle 
C  E' D  (since  E'  D  is  the  base,  and  C'  C  the  altitude 
of  this  triangle).     One  side,  CD,  remaining  constant, 
we  can  represent  the  area  of  the  triangle  by  the  altitude 
E'  ef ,  or  the  torque  is  proportional  to  the  abcissa   of 
the  point  Ef. 

Now,  there  are  a  number  of  constant  losses,  such  as 
those  due  to  bearing  and  air  friction,  iron  losses  and 
copper  losses  at  no  load.  These  losses  being  constant, 
manifest  themselves  in  the  diagram  by  a  constant 
diminution  of  the  torque.  In  other  words,  the  line  rep- 
resenting the  torque,  E'  e' ,  is  for  example,  diminished 
by  a  length  e1  e^,  proportional  to  the  constant  losses. 
Therefore,  drawing  a  line  parallel  to  the  axis  of  F,  and 


14  THE  INDUCTION  MOTOR. 

at  a  distance  from  it  equal  to  e'  e/,  we  find  the  actual 
torque  represented  by  the  line  e^'  E',  or 

Torque  oc^/  Ef 

(c)  The   output  is  given  by  considering  the   loss  in 
the    short-circuited    secondary    E'  F'    as    the    reduced 
altitude  of  the  triangle  C  Ff  D,  or 

Output  oc  //  F' 

(d)  The  slip  results  from  the  fact  that  the  currents 
produced    in    the    secondary    are    proportional    to    the 
product  of  secondary  field  and  the  slip,  or 

•          secondary  current      C'  C 
Slip  oc  -   —  -=  -  cc  .. 

secondary  field       E'  D 


E'  C 
But  C'  C  oc  E'  C,  and  therefore,  also,  slip  oc  -rz 


Draw  in  the  diagram  any  straight  line  F/  C/,  so 
that  the  angle  DElfClf=  D  E'  C.  D  E'  C  is  con- 
stant for  all  loads,  since  all  angles  inscribed  in  the 
same  segment  are  equal,  and  we  have 

E'C  =  F/CV 
E'  D  ~  F/  D 

In  the  above  proportion  F/  D  remains  constant;  then 
E   C       _.,„., 


E'D 


Therefore,  the  slip  is  proportional  to  £/  C\'.  Thus 
we  have  in  Elf  C\'  a  measure  of  the  slip. 

In  order  to  find  the  value  of  the  slip,  we  may  deter- 
mine the  point  where  the  slip  is  100%,  that  is  when 


THE  INDUCTION  MOTOR,  15 

the  rotor  is  stationary.  In  this  case  the  output  must 
be  zero,  and  the  point  F  must  coincide  with  D.  This 
happens  when  C'  coincides  with  Ck  and  the  line  Ck  D 
is  tangent  to  the  circle  OF.  Draw  the  line  5  Ck  parallel 
to  E/C/;  then  the  intercept  s  s'  is  proportional  to 
E/C/,  and  is  in  direct  ratio  to  s  Ck  (100%  slip)  or 
slip  =  5  S'. 

Finally,  the  electrical  efficiency  may  be  determined. 

The  electrical  loss  in  per  cent,  of  the  input  is  pro- 
portional to 

secondary  current2     C'C2     C'  C  r  .          .„      „,„     ~/rYI 

GO-— ^-.oo  [since  c'C  oo  C'C  x  CD] 

Input  c'C'     C' DL 

Drawing  a  line  Ck  f  from  the  point  Ck  at  right  angles 
to  A  D,  we  obtain  directly  the  above  ratio  (as  in  the 
case  of  the  slip)  in  the  line  7-7-',  and  the  electrical  effi- 
ciency in  the  remainder  of  the  line  7-  Ck,  or 

Electrical  Efficiency  =  Ck  f. 

The  diagram  now  gives  us  a  complete  determination 
of  all  the  essential  characteristics  of  an  induction  motor. 

At  no  load  (Fig.  7),  the  point  C°  is  fully  determined 
by  the  no  load  losses,  viz.,  the  iron  losses,  friction  and 
windage,  and  the  copper  loss  of  the  magnetizing  cur- 
rent. These  losses  are  represented  by  an  energy  cur- 
rent C  C°,  which,  combined  with  the  magnetizing 
current  A  C  gives  the  no  load  current  A  C°. 

The  torque  and  output  are  zero,  and  the  slip  very 
small.  With  increasing  load,  the  point  C'  moves  along 
the  circumference  of  the  circle  Oc.  The  phase  difference 
6  between  the  electromotive  force  and  the  current  de- 
creases, that  is,  the  power  factor  increases.  At  the 
point  C',  where  A  C'  is  tangent  to  the  circle  Or,  the 


16 


THE  INDUCTION  MOTOR. 


phase  difference  is  a  minimum.     The  torque  is  e'  E', 
the  output  /'  F',  and  the  slip  5  S'. 

The  output  increases  with  increasing  slip,  and   reaches 


1) 


FIG.  7. 

a  maximum  value  at  the  point  F" .  If  the  slip  still 
further  increases,  the  output  diminishes;  the  input 
still  increases,  but  later  also  decreases,  while  the  phase 
difference  rapidly  increases.  The  output  becomes 


THE  INDUCTION  MOTOR.  17 

equal  to  zero  when  the  slip  equals  100%,  and  the  point 
5'  coincides  with  the  point  Ck.  The  total  input  is 
then  consumed  in  the  motor. 

PRACTICAL  APPLICATION  OF  THE  DIAGRAM. 

In  the  preceding  pages,  we  have  derived  fully  and 
definitely  all  characteristics  chiefly  from  theoretical 
considerations.  It  remains  to  show  the  agreement 
between  the  theory  and  practical  tests.  If  .we  have 
the  completed  motor  before  us,  we  need  the  following 
data  in  order  to  determine  its  properties. 

1.   Ratio  of  the 

Reluctance  of  leakage  path 


Reluctance  of  path  of  main  field. 

2.  Resistance  of  the  circuits. 

3.  Iron  and  friction  losses. 

4.  Current,  motor  running  light. 

5.  Current,  motor  blocked. 

Some  of  these  qualities,  as,  for  instance,  the  reluct- 
ances, cannot  be  directly  measured.  As  we  have  deter- 
mined in  the  preceding  work,  it  would  be  possible  to 
measure  these  reluctances  if  the  losses  were  negligible, 
that  is,  if  the  electrical  resistances  were  equal  to  zero. 
Then,  if  we  measure  the  primary  current,  with  the  sec- 
ondary circuit  open,  and  also  the  current  with  the 
secondary  circuit  closed  and  stationary,  we  would  have, 
in  the  first  case,  the  value  A  C,  and  in  the  second  case 
the  value  A  D  and 

A  D  _        reluctance  of  leakage  path 
A  C        reluctance  of  path  of  main   field 

As  shown,  in  consequence  of  the  losses,  the  actual, 
limiting  points  are  C°  and  Ck,  instead  of  C  and  D. 


18  THE  INDUCTION  MOTOR. 

These  points  are  easily  obtained,  and  from  them  the 
centre  of  the  circle  Oc.  We  then  find  the  points  C 
and  D,  and  all  other  characteristic  points  in  a  similar 
manner.  It  is  only  necessary  to  perform  two  tests 
with  ammeter  and  wattmeter. 

(1)  Motor  running  without  load.     The  current  A  C° 
and  the  angle  of  phase  difference  6  =  C°  A  B. 

(2)  With  stationary  and  short-circuited  rotor.     The 
current  A  Ck  and  the  angle  of  phase  difference  0  =  Ck  A  B. 

If  it  is  impossible  to  allow  full  short-circuit  current 
to  flow,  it  is  sufficient  to  make  the  test  at  a  lower  im- 
pressed electromotive  force.  The  real  short-circuit 
current  is  then  proportional  to  the  electromotive  force 
applied.  If  we  represent  the  electromotive  force  in 
direction  by  the  line  A  B;  the  direction  and  magnitude 
of  the  no  load  current  by  A  C°,  and  the  direction  and 
magnitude  of  the  short-circuit  current  by  A  Ck,  then 
the  center  Oc  of  the  circle  is  determined,  since  it  must 
lie  on  a  line  at  right  angles  to  A  B.  If,  by  a  third 
test,  the  resistance  R^  of  the  primary  circuit  is  deter- 
mined, the  point  Ek  can  be  found  by  making  Ck  Ek 
proportional  to  7\  R^  Thence  we  find  the  centers 
OB  and  OF  and  the  slip  line  5  Ck. 

These  two  simple  tests,  one  at  no  load,  and  one 
with  short-circuited  and  stationary  rotor,  give  sufficient 
data  to  indicate  clearly  all  characteristic  points,  and 
to  construct  the  load  curves.  Since  the  only  tests 
necessary  are  on  the  motor  running  light,  and  with 
the  rotor  blocked,  the  method  renders  any  exact  load 
tests  unnecessary.  Moreover,  the  diagram  gives  at  a 
glance  the  reasons  for  the  various  characteristics  of 
the  motor,  and  shows  how  the  proportions  need  to  be 
changed  to  obtain  any  desired  result. 

For  instance,  in  a  motor  with  low  electrical  losses, 


THE  INDUCTION  MOTOR.  19 

the  line  Ek  D  which  represents  them  will  be  very 
small.  Therefore,  the  point  Ek  will  be  very  high,  and 
the  starting  torque  Ek  ek  will  be  small.  As  is  well 
known,  this  may  be  avoided  by  inserting,  by  means 
of  slip  rings,  resistances  in  the  secondary  circuits. 

If  it  is  desired  to  avoid  slip  rings,  the  resistance  of 
the  short-circuited  secondary  must  not  be  made  too 
small.  Then  the  points  Ek  and  Ck  are  removed  farther 
from  D.  Moreover,  the  variation  of  the  slip  line  5  Ck 
and  the  slip  is  greater. 

The  motor  will  have,  under  load,  greater  speed 
variations  and  low  efficiency. 

It  is  possible,  however,  to  obtain  good  torque  without 
excessive  losses,  if  the  motor  is  so  designed  that  the 
normal  load  is  reached  at  a  point  far  beyond  Cf,  that 
is,  if  the  field  is  made  strong.  But  in  this  case  the 
magnetizing  current  is  very  large  in  proportion  to  the 
energy  current  and  cos  6  will  be  unfavorably  small. 
Such  a  motor  will,  moreover,  be  comparatively  large. 
The  size  of  the  motor  is,  in  general,  determined  by  the 
maximum  abscissa  Oc  C"  ' ,  The  abscissae  for  normal 
load  c'  C'  is  then  chosen  so  that  the  desired  relation 
between  normal  load  and  maximum  load  is  obtained. 
The  leakage  constant  of  Rothert,  which  fixes  for  each 
overload  capacity  a  certain  ratio  between  current  and 
field,  corresponds  with  this  point  of  view. 

In  the  diagram  this  relation  is  expressed  by  the  ratio 

,       ,  that  is  by  the  tangent  of  the  angle  C'  D  C.     If 

this  angle  becomes  45°,  the  motor  has  its  maximum  energy 
input.  The  smaller  this  angle,  the  larger  the  maximum 
capacity  of  the  motor.  Rothert  does  not  consider  in 
his  constant  the  magnetizing  current.  The  overload 
capacity  of  a  certain  motor  type  is  limited  by  the 


20  THE  INDUCTION  MOTOR. 

magnetizing  current,  which  increases  so  that  the  phase 
difference  becomes  large  and  the  power  factor  small. 

The  magnetizing  current  is  determined  by  the  air- 
gap.  If  the  air-gap  is  changed,  the  magnetizing  current, 
A  C,  changes;  while  the  diagram  above  the  point  C 
regains  unchanged. 

Therefore,  we  can  say  with  Rothert  that,  disregarding 
the  magnetizing  current,  the  diagrams  for  all  multi- 
phase motors  are  identical. 

The  maximum  input  C'"  is  proportional  to  the  energy 
current,  that  is,  to  the  radius  of  the  circle.  In  case  of 
double  input  capacity,  for  instance,  the  normal  input 
corresponds  to  one  half  of  the  radius. 

It  is  possible  to  judge  of  the  output  of  a  motor  ap- 
proximately by  the  short  circuit  current.  Since  the 
latter  is  nearly  'equal  to  the  diameter  of  the  circle,  we 
may  employ  the  following  rough  approximation: 

The  normal  capacity  of  a  motor  corresponds  to  about 
one-quarter  of  its  short  circuit  current,  that  is,  if  a 
motor  of  100  volts  has  a  short-circuit  current  of  100 
amperes  its  normal  output  corresponds  to 

iXlOOXlOOX\/3  =  4300   watts,    or   about    5    h.p., 

assuming  75%  efficiency,  that  is,  the  normal  output 
corresponds  to  one  quarter  of  the  current,  and  is  pro- 
portional to  one  quarter  of  the  current  necessary  to 
force  the  total  field  through  the  air-gap. 

Obviously,  with  constant  impressed  electromotive 
force,  this  current  varies  inversely  as  the  square  of  the 
turns,  directly  with  the  length  of  the  leakage  path, 
and  therefore  with  the  diameter  d,  and  inversely  with 
the  width  of  the  effective  iron;  hence 


~j  X  ~n 

b          m2 


THE  INDUCTION  MOTOR.  21 

For  large  output,  therefore,  d  must  be  large  and  b 
small.* 

These  interesting,  and  extremely  simple,  considera- 
tions become,  after  a  little  practice,  very  serviceable. 

Considering  that  w  can  be  replaced  by 

electromotive  force 


and  L  by  electromotive  force    X   current  i,  we  obtain 
all  values  of  the  Rothert  leakage  constant,  or 

mi  b 


Good  design  in  standard  motors  requires  good  effi- 
ciency and  low  heating,  and  should  further  fulfil  the 
following  conditions:  maximum  output  of  one  and 
one-half  to  two  times  the  normal  output  ;  starting 
torque  about  equal  to  normal  full  load  running  torque 
and  moderate  starting  current;  small  "no  load"  cur- 
rent, varying  from  one-third  to  one-fifth  of  the  normal 
full  load  current,  depending  on  the  size  of  the  motor; 
the  power  factor  should  be  high  and  the  slip  small. 

The  above  conditions  being  more  or  less  conflicting, 

*  This  statement  does  not  agree  with  the  results  of  modern  re- 
search made  by  Behn-Eschenburg  and  others.  It  has  been 
shown  and  generally  recognized  that  the  end  connection  leakage 
is  a  considerable  part  of  the  total  leakage,  and  that  this  kind  of 
leakage  is  large  in  motors  which  have  a  comparatively  large  di- 
ameter and  a  small  iron  width  ;  therefore  the  output  of  a  motor 
is  not  always  improved  by  simply  making  d  larger  and  b  smaller. 
In  very  many  cases,  the  opposite  effect  is  obtained.  (See  Jour- 
nal of  the  Institute  of  Electrical  Engineers,  Vol.  33,  page  239, 
April,  1904.)  Paper  by  Behn-Eschenburg  on  Magnetic  Disper- 
sion in  Induction  Motors.  —  ROWE-HELLMUND. 


22 


THE  INDUCTION  MOTOR. 


it   is  not   always   possible   to   satisfy   them   all   at  the 
same  time,  especially  in  small  motors. 

The  derivation  of  the  circle  diagram  above  given,  contains  a 
simplification  which  leads  to  a  slight  inaccuracy  which  did  not 
appear  in  the  original  derivation,  published  in  the  Electrotech- 
msche  Zeitschrift,  October  11,  1894,  to  which  reference  has  al- 
ready been  made. 

In  consequence  of  the  resistance  of  the  primary  winding,  the 


FIG.  8. 


theoretical  magnetizing  current  is  not  entirely  wattless,  and  its 
phase  difference,  with  respect  to  the  primary  impressed  electro- 
motive force,  is  somewhat  less  than  90°.  Thus,  the  magnet- 
izing current  A  C  is  not  exactly  at  right  angles  to  A  B,  and  the 
center  of  the  circle  Oc  is  not  exactly  in  the  perpendicular  erected 
at  A  on  A  B.  Practically,  the  inaccuracy  introduced  is  so 
small  that  in  most  cases  it  does  not  introduce  a  measurable 
error.  In  very  small  motors,  only,  is  the  result  of  the  in- 


THE  INDUCTION  MOTOR.  23 

accuracy  noticeable,  and  in  these  cases,  the  result  is  always 
somewhat  better  than  that  shown  by  the  diagrams. 

In  the  first  rough  laying  out  of  a  motor,  this  refinement  will 
be  found  unnecessary,  inasmuch  as  the  use  of  the  uncorrected 
diagram  introduces  only  a  factor  of  safety,  while  in  cases  of 
relatively  high  stator  resistance  the  correction  is  easily  made. 

For  the  sake  of  completeness,  the  method  of  applying  the 
correction  will  be  given:  The  original  circle  diagram  on  page 
563,  E.T.Z.,  1894,  had  the  form  shown  in  Fig.  8.  The  influ- 
ence of  the  primary  resistance,  in  causing  a  shifting  of  the  line 
A  D',  representing  the  magnetizing  current,  has  been  much 
exaggerated  in  the  figure. 

The  theoretically  exact  position  for  the  center  of  the  circle 
Oc  is  determined  as  follows:  The  point  C  for  the  theoretical  no 
load  condition  (assuming  no  friction  and  no  core  losses),  as 
well  as  the  point  CIv  for  the  theoretical  short  circuit  condition 
(the  resistance  of  secondary  being  equal  to  zero),  must  be 
situated  on  a  half  circle,  the  diameter  of  which  is  the  impressed 
electromotive  force  A  B.  The  center  of  the  circle  Oc  is  on  the 
intersection  of  the  tangent  to  the  semi-circle  A  B  at  C  and  CIv. 

The  center  of  the  circle  Oc  is  not,  therefore,  exactly  on  the 
perpendicular  to  A  B,  but  on  a  line  inclined  to  it  by  an  angle  e. 
This  angle  is 

<  s  =  2  C  B  A,  and  sin  C  B  A  =  dj 

In  motors,  as  usually  constructed,  this  ratio  representing  the 
relation  between  the  drop  of  potential  caused  by  the  magnet- 
izing current  and  the  impressed  electromotive  force,  is  so  small 
that  the  result  is  not  influenced  by  a  measurable  amount. 
Therefore,  we  shall  use  the  approximate  diagram  in  the  fol- 
lowing examples  illustrating  the  application  of  the  diagram. 

APPLICATION   OF   THE   DIAGRAM. 

It  may  be  well  to  exemplify  the  preceding  considera- 
tions by  applying  them  to  several  practical  cases.  The 
motors  considered  in  the  following  were  built  by  the 
"  Societe  Electricite  et  Hydraulique  "  and  were  in- 
tended to  give  a  starting  torque  equal  to  twice  the 
full  load  running  torque. 


24  THE  INDUCTION  MOTOR. 

Two  Horse  Power  Motor'. 

1.  At  no  load,  120  volts  applied. 

4.1  amperes, 
210  watts, 

2.  Short-circuited  secondary  and  stationary  rotor,  120 
v.  applied. 

55  amperes, 
6940  watts, 

3.  Resistance  of  primary  member  per  phase. 

.344  ohm. 

From  the  foregoing  data,  the  diagram  (Fig.  9)  was 
constructed. 

From  the  watts  at  no  load,  we  obtain  the  energy 
current : 

210 

7"  =  TW3  ==  ]  ampere' 

First  construct  the  no  load  current  triangle  A  C  C°, 
in  which 

A  C°  =  no  load  current  =4.1  amperes 

C  C°  =  energy  current  =  1 

A  C    =  magnetizing  current  =  4 

The  energy  current  C  C°  represents  the  no  load  losses 
by  friction,  hysteresis  and  eddy  currents.  These  losses 
are  assumed  to  be  constant  for  all  loads.  Therefore, 
they  must  be  allowed  for  by  subtracting  a  constant 
amount  from  the  mechanical  output  and  torque.  This 
can  be  accomplished  in  the  diagram  by  drawing  a  line 
x  y  through  C°  parallel  to  A  C  and  measuring  the  output 
and  torque  from  the  line  x  y.  At  the  point  C',  for 
instance,  the  input  is  Cf  cf ,  the  torque  Ef  £/,  and  the 
output  //F'. 


THE  INDUCTION  MOTOR. 


25 


120  Volts 


26  THE  INDUCTION  MOTOR. 

From  the  stationary  test  we  obtain  an  energy  current 
6940 


120V3 


33.5  amperes,  01 


a       33.5         _ 

cos  6  =  — —  =  .61 
oo 


This  value  enables  us  to  determine  the  point  Ck, 
since  A  Ck  is  proportional  to  the  current  with  rotor 
blocked,  or  55  amperes,  and  cos  Ck  A  R  equals  the  power 
factor,  or  .61. 

From  these  points  C°  and  Ck,  it  is  easy  to  find  the 
center  Oc,  and  the  corresponding  circle  C  C°  Ck  D. 

If  now  the  line  Ck  D  is  drawn,  then  the  circle  giving 
the  mechanical  output  is  determined,  since  we  know 
that  the  output  of  the  short  circuited  and  stationary 
rotor  corresponding  to  the  point  Ck  is  zero.  Therefore, 
the  intersection  of  the  line  Ck  D  with  the  circle  must 
fall  at  the  point  D.  The  line  Ck  D  must  therefore  be 
tangent  to  the  output  circle  at  D.  Hence,  the  center 
0E-and  the  circle  E'  itself  is  determined. 

As  we  have  seen,  the  mechanical  output  derived 
from  this  circle  must  be. decreased  by  an  amount  equal 
to  the  no  load  losses  which  are  represented  by  the 
parallel  line  xy,  or,  in  other  words,  the  output  is  E'  ev 
etc. 

Finally,  in  order  to  find  the  third  circle  for  the  torque, 
we  must  remember  that  the  line  Ck  D  represents  the 
electrical  losses  of  locked  rotor.  These  losses  divide 
themselves  into  two  parts,  primary  losses  and  secondary 
losses,  which  are  proportional,  respectively,  to  the 
reduced  resistance  in  each  member. 

Since  the  total  field  A  D  corresponds  to  an  electro- 


THE  INDUCTION  MOTOR. 


27 


motive  force  of  120  volts,  we  find  in  the  same  scale  the 
loss  vector  at  short  circuit. 


CkD  X 


=  73  volts. 


In  the  primary  member,  of  which  the  resistance  per 
phase  is  .344  ohm,  and  in  which  the  short  circuit  current 
is  55  amperes,  the  drop  of  potential  is 

V3X55X.344  =  32.7  volts. 


4  H.P. 


If  we  therefore  make  Ck  Ek  equal  to  Ck  D  multiplied 

oo    1-7 

by  '—=^-,   we   know  that   the   torque   circle   must   pass 
7o 

through  the  point  Ek,  and  therefore  this  'circle,  and  its 
center  OE  are  determined. 

Finally,  it  remains  to  draw  the  line  Ck  s  from  Ck  at 
right  angles  to  0E  D,  from  which  may  be  found  the  slip 
for  any  load.  At  no  load  the  slip  is  zero,  and  at  full 
load  5%. 

The  diagram  now  shows  all  the  characteristic  proper- 
ties of  the  motor.  It  is  possible  to  transfer,  by  means 


28  THE  INDUCTION  MOTOR. 

of  a  compass,  all  of  the  various  values  directly  to  a 
rectangular  system,  as  shown  in  Fig.  10. 

The  normal  output  of  the  motor  is  two  horse  power 
at  78%  efficiency,  86%  power  factor  and  5%  slip. 
The  maximum  output  is  more  than  double  the  normal 
full  load  value.  The  maximum  power  factor  is  .90. 
Finally,  the  distance  between  Ck  and  the  line  X  Y  gives 


FIG.  11. 

a  starting  torque  of  four  synchronous  horse  power,  the 
starting  current  being  55  amperes.  The  motor  thus 
meets  the  requirements  usually  assumed. 

The  above  motor  is  a  small  one ;  in  the  diagram  for  which 
the  center  Oc  should  be  shifted  from  the  line  at  right  angles  to 
A  B,  as  was  shown  in  the  note  on  page  16,  it  may  be  here  shown 
that  this  correction  is  very  small  and  that  it  may  be  practically 
neglected. 


THE  INDUCTION  MOTOR.  29 

We  have: 

Impressed  electromotive  force  ..............  120  volts 

Magnetizing  current  ......................  .      4  amperes 

Resistance  of  stator  winding  per  phase  .......         344  ohm 

and  therefore 


The  angle  e  is  shown  in  Fig.  11,  and  the  dotted  circle  shows 
the  corrected  position  of  the  circle. 

If  one  calculates  results  from  this  corrected  circle,  it  will  be 
seen  that  the  maximum  power  factor  has  increased  from  .  9 
to  .91,  and  that  other  values,  such  as  efficiency  and  maximum 
output,  are  unchanged.  The  difference  in  power  factor  is 
smaller  than  can  be  observed  with  ordinary  measuring  instru- 
ments. 

As  a  matter  of  fact,  only  in  very  small  low  speed  motors  is  a 
difference  noticeable,  and  in  these  cases  the  correction  may 
be  made.  Exact  calculations  are  anyhow  difficult  in  the  case 
of  small  motors,  and  since  we  know  that  the  actual  values  are 
somewhat  better  than  those  given  by  the  diagram,  it  is  hardly 
necessary  to  make  a  correction. 

SEVEN   HORSE   POWER  MOTOR. 

Tests  on  a  seven  horse  power  motor  gave  the  fol- 
lowing data: 

1.  At  no  load,  110  volts  per  phase. 
/  =  6.3  amperes  per  phase, 

A  =  100  watts  per  phase. 

2.  Locked   and   short-circuited   rotor,    110   volts   per 
phase. 

/  =  118  amperes  per  phase, 
A  =  6670  watts  per  phase. 

3.  Resistance  per  phase  =  .142  ohm. 

From  these  results,  the  diagram  shown  in  Fig.  12 
was  constructed. 


30 


THE  INDUCTION  MOTOR. 


The  no  load  energy  current  is 

Iw  =  r—    =  -91  ampere. 


110  Volts 


Output 

Torque 

Electrical  Input  c 
C 

A 


I'O        20        30        40    '     50        60 
<-0.9  Amp. 

FIG.  12. 


Draw  the  current  triangle  A  C  C°,  in  which 
A  C°  —  no  load  current  =  6.3  amperes 
.  C  C°  =  energy  current     =     .91     " 
A  C    =  magnetizing  current  =  6.2  amperes. 


THE  INDUCTION  MOTOR.  31 

For  the  no  load  losses,  draw  the  line  H  Y,  parallel 
to  A  C,  passing  through  the  point  C°.  The  mechanical 
output,  as  we  know,  is  measured  from  the  line  h  y, 
since  the  constant  no  load  losses  have  always  to  be 
subtracted.  The  test  of  the  stationary  and  short- 
circuited  rotor,  gives  us  the  energy  current, 

—  -rr-  =  60.7  amperes  =  i  cos  6  or 

60.7 
cos  6  =  =  .515  power  factor. 

llo 

Thus  we  find  the  point  Ck  of  the  diagram.  The  center 
OE  of  the  input  circle  can  now  be  easily  found  by  con- 
struction. The  point  OP  is  obtained  by  drawing  through 
D  a  perpendicular  to  Ck  D  until  it  intersects  the  line 
OE  OF  in  the  desired  point  OF.  In  other  words,  the  line 
D  Ck  is  a  tangent  to  the  circle  of  mechanical  output. 
It  now  remains  to  find  the  third,  or  torque  circle. 
This  is  done  by  dividing  the  line  Ck  D  which  represents 
the  electrical  losses  at  short  circuit  in  the  ratio  of  the 
resistances  of  the  primary  and  secondary  circuits. 
The  third  circle  then  passes  through  this  point. 

In  the  diagram,  the  field  per  phase  corresponds  to  a 
potential  of  110  volts,  and  the  loss  vector  at  short 
circuit,  measured  in  the  same  scale,  is 

C*DX110 


If  the  primary  winding,  which  has  a  resistance  of  .142 
ohms  per  phase,  we  have  118  amperes  at  short  circuit, 
therefore  the  drop  of  potential  is  118  X  .142  =  16.8  volts. 

To  find  the  intersection  of  the  third  circle  with  Ck  D, 


32 


THE  INDUCTION  MOTOR. 


10  H.P. 


D  no  Volts 


Torque     \\^-^/-- 
Electrical  Input  c^-^^^ 

A 


10     20    30    40    50    60    70    80    90   100 

FIG.  14. 


THE  INDUCTION  MOTOR. 


33 


we  must  lay  off  this  value  =  Ck  Ek  on  Ck  D  in  the  ratio 
of  16.8  to  57. 

Finally,  erect  a  perpendicular  from  Ck  on  the  line 
OE  D  and  obtain  the  slip  for  any  load.  At  no  load  the 
slip  is  very  small;  at  full  load  only  6.6%. 
.  Fig.  13  shows  the  characteristic  curves  of  the  motor 
as  functions  of  the  output.  The  full  load  output  is 
seven  horse  power.  At  this  load,  the  efficiency  is 


8       9       10      11      12      13      14      15  H.P 


85.5%,  and  the  power  factor  90%.  The  power  factor 
reaches  a  maximum  of  about  91%  at  a  load  of  about 
nine  horse  power.  The  motor  starts  with  a  torque 
of  16.7  synchronous  horse-power,  and  has  a  maximum 
output  of  15.4  horse-power. 

TEN  HORSE  POWER  MOTOR. 

Figs.    14  and    15   give   the   phase   diagram   and   the 
characteristic  curves  of  a  10  horse  power  motor. 


34  THE  INDUCTION  MOTOR. 

The  tests  gave  the  following  results: 

1.  At  no  load,  110  volts  applied  per  phase. 
1  =  8  amperes, 

A  =  197  watts. 

2.  Rotor  locked  and  short-circuited. 
110  volts  applied. 

/  =  162.5  amperes, 
A  =  9  kw. 

3.  Resistance  per  phase  =  .123  ohm. 

Fig.  14  is  the  diagram  constructed  from  these  data. 
Fig.  15  shows  the  characteristic  curves.  The  following 
results  are  obtained  at  full  load  of  10  horse  power: 

Efficiency  =  83% 

Current  =  29.5  amperes 

Electrical  Input   =  2.95  kw. 

Power  Factor       =  .915 

Slip  =  6% 

Maximum  output  =  21  horse  power. 

THE    INDUCTION   MOTOR   AS  GENERATOR. 

It  is  a  well  known  fact  that  an  induction  motor,  if 
driven  above  synchronism,  will  act  as  a  generator 
analagously  to  the  direct  current  shunt  machine.  It 
is  interesting  that  the  circle  diagram  gives  also  light 
on  this  use  of  the  motor,  if  the  circles  are  completed 
on  the  left  hand  side  of  the  line  A  D  as  in  Fig.  16.  The 
slip  is  now  negative,  and  what  was  before  mechanical 
output,  is  now  mechanical  input.  Further,  the  elec- 
trical input  now  becomes  the  electrical  output. 

We  see  that  the  generator  gives  now  exactly  the  same 
current  for  the  same  phase  difference,  as  when  employed 
as  a  motor.  In  other  words,  the  electrical  energy  is 
the  same  in  both  cases.  The  mechanical  input  is,  of 
course,  considerably  larger  in  the  case  of  the  gen- 


THE  INDUCTION  MOTOR. 


35 


FIG.  16. 


3          Input         4  H.P. 


FlG.   17. 


36  THE  INDUCTION  MOTOR. 

erator  than  the  mechanical  output  of  the  motor.  It 
is  easily  seen  that  the  difference  amounts  to  twico 
the  losses,  since  now  all  losses  add  themselves  to  the. 
electrical  output,  while  in  the  motor  they  were  sub- 
tracted from  the  electrical  input. 

From  a  theoretical  standpoint,  the  curves  become 
exceedingly  interesting  if  drawn,  not  as  functions  to 
the  mechanical  output,  but  as  functions  to  the  elec- 
trical input;  that  is,  as  functions  of  the  values  which 
correspond  to  the  energy  currents  which  are  the  ab- 
scissae of  the  diagram  (Fig.  18). 

The  curves  thus  obtained  are  of  regular  mathematical 
forms,  and  are  simple  closed  curves.  The  curves  of 
torque  and  mechanical  output  are  ellipses.  The  power 
factor  curve  is  a  function  of  an  angle,  etc. 

There  are  three  essentially  different  parts  of  the 
curves  to  be  considered: 

1.  As  a  motor,  (slip  0  to  1)  C0 — >Ck. 

2.  As  a  generator    j  slip  <  0  )   „  _^_^k 

3.  As  a  generator    (  slip  >  1  I 

With  these  general  diagrams  and  curves,  all  essential 
points  of  the  use  of  the  present  method  are  exhausted. 
All  complicated  details  of  the  motor  are  based  on  two 
phenomena,  magnetic  leakage  and  electric  resistance. 
These  are  essentially  the  only  values  which  determine 
the  operation  of  the  motor,  and  it  remains  for  the 
designer  to  choose  the  proper  proportions. 

SINGLE    PHASE   MOTORS. 

The  fundamental  theory  of  the  polyphase  motor  is 
much  simplified  by  the  introduction  of  the  principle 
of  the  rotating  field,  but  the  theory  of  the  single-phase 
motor  is  somewhat  more  difficult. 


THE  INDUCTION  MOTOR. 


37 


38 


THE  INDUCTION  MOTOR. 


The  single-phase  motor  possesses  a  single  phase 
winding  for  a  simple  alternating  current  (Fig.  19),  and 
can  therefore  produce  a  simple  alternating  field  in  the 
direction  of  the  coil  axis  only.  Such  a  field,  of  course, 
will  not  cause  rotation  of  a  short-circuited  armature, 
without  commutator  or  similar  devices.  Therefore  it 
seems,  at  first  thought,  somewhat  astonishing,  that 
such  a  motor  can  operate  at  all,  and  indeed,  in  its 


FIG.  19. 


simplest  form  such  a  motor  will  not  start;  however, 
after  the  motor  has  been  started  by  any  means  what- 
soever, it  then  operates  exactly  like  a  polyphase  motor. 
This  phenomenon  is  explained  by  the  fact  that  in  the 
single  phase  motor,  rotating  at  a  certain  speed,  there 
is  a  field  caused  in  the  secondary  which  rotates  with  it. 
Between  this  field  and  the  ampere  turns  of  the  single 
phase  winding,  a  torque  is  established  in  the  same 
manner  as  between  the  field  of  a  short-circuited  second- 


THE  INDUCTION  MOTOR.  39 

ary  and  one  of  the  two  or  three  phase  windings  of 
the  polyphase  motor.  The  torque,  however,  is  pulsat- 
ing, since  there  is  no  second  phase,  and  it  becomes  a 
maximum,  falls  to  zero  and  again  reaches  a  maximum 
varying  with  the  alternations  of  the  field. 

In  the  case  of  induction  motors,  single  as  well  as  poly- 
phase, we  may  consider  the  short-circuited  secondary 
as  a  field  magnet  of  a  synchronous  motor.  The  only 
difference  is  that  the  field  does  not  rotate  in  the  in- 
duction motor  in  synchronism  with  the  armature,  but 
has  a  small  relative  rotation  with  respect  to  the  arma- 
ture, corresponding  to  the  slip. 

In  the  short-circuited  secondary  of  a  single-phase 
motor  there  exists,  just  as  in  the  polyphase  motor,  a 
constant  rotating  field  notwithstanding  the  fact  that 
the  current  exciting  winding  produces  only  a  simple 
alternating  field.  This  is  explained  by  the  character- 
istic property  of  the  short-circuited  winding.  We 
know  that  any  short-circuited  winding  tends  strongly  to 
keep  up  the  field  interlinked  with  it,  since  any  change 
in  this  field  causes  current  in  the  short-circuited  wind- 
ing which  opposes  such  change.  This  counteracting 
effect  is  the  larger  the  lower  the  resistance  of  the  wind- 
ing and  the  greater  the  frequency. 

If  we  consider  the  secondary  of  a  single  phase  motor 
revolving  synchronously,  the  primary  alternating  field 
will  not  appear  as  an  alternating  field  in  the  secondary, 
since  in  the  same  time,  during  which  the  field  intensity 
starting  from  zero  reaches  a  maximum  and  decreases 
again  to  zero,  the  secondary  has  turned  through  180°, 
and  the  primary  alternating  field  will  therefore  appear 
as  a  pulsating  field,  and  will  have  always  the  same 
direction  in  regard  to  the  secondary  member. 

Each  pulsation  of  the  secondary  field  causes,   how- 


40 


THE  INDUCTION  MOTOR. 


ever,  in  the  secondary  winding  wattless  currents  which 
diminish  the  pulsation  to  a  negligible  minimum,  i.e,, 
they  keep  the  pulsating  field  of  the  short-circuited 
secondary  practically  constant.  In  the  positions  where 
the  field  corresponding  to  the  exciting  current  tends 
to  exceed  this  constant  value,  the  secondary  ampere 
turns  demagnetize.  In  other  positions  they  furnish 
the  magnetizing  current. 


FIG.  20. 


FIG.  22. 


FIG.  23. 


\ 


FIG.  24. 


In  order  to  obtain  a  still  clearer  understanding  of 
this  phenomena,  let  us  consider  the  four  characteristic 
positions: 

In  Figs.  20-23,  let  a  b  be  the  primary  winding,  and 
c  the  secondary,  which  rotates  synchronously. 

The  field  caused  by  the  current  7,  will  then  be  as 
shown  in  Fig  24. 


THE  INDUCTION  MOTOR. 


41 


In  the  positions  1  and  3,  the  maximum  value  of  the 
current  acts  once  in  one  direction  and  once  in  the 
other.  Since,  however,  the  rotor  has  meanwhile  turned 
through  180°,  its  field  will  have  in  the  positions  1  and  3 
the  same  direction,  and  will  have  a  characteristic 
shown  by  the  dotted  line.  The  above  described  action 
of  the  rotor  now  appears,  and  prevents  the  field  from 


FIG.  25. 


being  strongly  pulsating,  the  pulsations  being  smoothed 
out.  Therefore,  we  obtain  a  constant  field  rotating 
with  the  rotor. 

It  is  now  possible  to  picture  the  action  of  the  rotor 
as  follows:  Let  7t  be  the  magnetizing  ampere  turns  of 
the  exciting  member,  /2  those  of  the  rotor;  then  these 
two  obviously  must  combine  to  form  the  ampere  turns 


42  THE  INDUCTION  MOTOR.  . 

1  1  which  produce  the  constant  field.     This  is   true,  if, 
for  the  moment,  we  neglect  the  leakage. 

If  we  assume  again  that  the  alternating  current  is  a 
sine  function,  and  that  7,  is  the  maximum  value  of  this 
current,  the  motor  being  unloaded,  then  the  instan- 
taneous value  of  the  ampere  turns  of  the  exciting 
member  at  no  load  is  proportional  to 

/!  oc  /t  sin  a 

If  we  consider  the  short-circuited  secondary  to  be 
stationary,  and  the  exciting  winding  to  rotate,  then 
the  constant  field  must  also  be  stationary,  This  field 
is  produced  by  the  ampere  turns  of  the  rotating  exciting 
member,  and  these  ampere  turns  vary  according  to  a 
sine  law.  Therefore,  the  field  of  the  short-circuited 
secondary  must  have  a  component  proportional  to 

T     .  7-2         7    1  ~~  2  cos  a 

/jSinaoc  T^sm2  a:  oc  7t   -    —  -  - 

The  field  of  the  short-circuited  secondary  must, 
however,  be  constant,  and  therefore  the  resulting 
ampere  turns  must  be  proportional  to  /,  where  /  is  the 
magnetizing  current.  The  difference  between  the  mag- 
netizing ampere  turns  and  those  active  in  the  exciting 
member  must,  therefore,  be  produced  in  the  short-cir- 
cuited secondary. 

Thus,  if  /2  is  the  value  of  the  current  in  the  secondary 
we  have 

/2  =  7-        (1  -cos  2  a) 


-  *-k  4  ~  cos  2  a 


THE  INDUCTION  MOTOR.  «3 

The  currents  in  the  secondary  can  only  be  periodic 
functions  of  the  time,  that  is,  there  can  be  no  constant 
terms.  Thus,  we  have, 


72  — ~  cos  2  a 

Therefore,  the  ampere  turns  of  the  primary  member 
are — 

A  Tl  oc  —-  — i  cos  2  a 
and  those  of  the  secondary  are — 

A  T2oc  ~ •  cos  2  a, 
and  the  magnetizing  ampere  turns 

These  equations  indicate  that  in  the  single-phase 
motor  the  magnetizing  current  /  is  half  of  the  no  load 
current.  The  four  characteristic  positions  are  now 
represented  in  Fig.  26. 

The  sum  of  the  two  fields  gives  a  straight  line  corre- 
sponding to  the  amplitude  ~. 


44  THE  INDUCTION  MOTOR. 

The  single  fields  are  represented  by  the  full  and 
dotted  waves.  The  rotating  field  is  not  quite  con- 
stant, but  is  slightly  pulsating.  It  is  obvious,  how- 
ever, that  this  may  be  neglected. 

Another  approximation  previously  made  will  now  be 
corrected.  Owing  to  the  presence  of  leakage,  the  ratio 

No  load  current 

is  not  exactlv  two. 


Magnetizing  current 

In   the   position   of   Figs.    27  and    28  the   secondary 
fields  are  the  same.     In  Fig.  27  it  is  induced  by  the 


>  sin  a  oc  3i  sin  2ar  I»  (Z-cos  5 


primary  winding  by  the  magnetizing  current  I  =  I  ^  —  72, 
or,  introducing  the  reluctance  of  the  secondary 


In  the  other  positions  (Fig.  28),  the  secondary  field 
is  produced  by  the  ampere  turns  of  the  short-circuited 
secondary.  The  reluctance  will  be  smaller,  since  the 
field  now  has  two  paths,  the  main  path  and  the  leakage 
path.  If  we  call  the  reluctance  in  this  case  pf  then 


THE  INDUCTION  MOTOR. 


45 


FIG.  28. 


46  THE  INDUCTION  MOTOR. 

We  have  seen  that  the  secondary  field  remains  almost 
constant,  and  therefore  we  have 


—  =  —  ,  or,  since 
P         P 


-j,y 

No  load  current 
Magnetizing  current 

By  substitution  of  the  simple  reluctances  p^  ps  and  p2 
the  ratio  can  appear  in  a  somewhat  different  form,  in 
which  only  known  values  appear.  It  will  be  seen  then 
that  p'  is  little  different  from  p  and  that  the  ratio  is 
almost  two. 

The  phenomena  in  single-phase  motors  are  thus  as 
follows: 

In  the  unloaded  single-phase  motor,  there  is  formed 
in  the  short-circuited  secondary,  as  in  the  polyphase 
motor,  a  field  which  rotates  with  the  short-circuited 
secondary.  This  field  when  it  coincides  with  the  axis 
of  the  primary  winding  is  produced  by  the  latter. 
At  right  angles  to  this  position,  it  is  produced  by  the 
no  load  currents  of  the  short-circuited  secondary.  The 
latter  are  wattless  and  are  produced  by  the  primary 
member.  Therefore,  the  no  load  current  equals  twice 
the  magnetizing  current,  or,  more  accurately,  the  no 

load  current  is  - — —  times  the  magnetizing  current. 


THE  INDUCTION  MOTOR.  47 

Obviously,  it  is  the  same  as  far  as  the  primary  winding 
is  concerned,  whether  the  field  in  the  position  at  right 
angles  to  the  primary  winding  is  produced  by  current 
in  the  short-circuited  secondary,  or  by  current  in  the 
winding  of  a  second-phase,  as  in  a  two-phase  motor. 

Therefore,  the  single  phase  primary  acts  with  the 
same  torque  on  a  short-circuited  secondary  as  that  of 
one-phase  of  a  two-phase  motor.  From  these  reflec- 
tions, we  are  able  to  obtain  the  diagram  for  the  single- 
phase  motor  from  that  of  the  polyphase. 

In  the  polyphase  motor  we  have 

A  C  oc  /  =  magnetizing  current. 
In  the  single  phase  motor  we  have 
A  C  =  / 


or  since  p  is  little  different  from  p' 
AC  =  21 

The  ratio-       ^  is  therefore   no   longer   -r— p  =  — ,   as 
/L    u  A   \s        p 

in  the  polyphase  motor,  but 

A_D  =  ^ 

A   C  ~  2/> 

This  ratio  is  equal  to  that  of  a  polyphase  motor 
having  double  the  reluctance. 

One  must,  however,  not  carry  the  analogy  too  far. 
For  instance,  we  must  not  expect  to  obtain  a  good 
single-phase  motor  by  running  a  polyphase  motor  on  a 
single-phase  circuit  by  simply  not  using  the  second-phase. 
Such  a  motor  would  give  bad  results,  and,  on  account 


48  THE  INDUCTION  MOTOR. 

of  the  omission  of  the  second  phase,  the  output  would 
not  much  exceed  one-half  the  output  of  the  polyphase 
motor  of  the  same  dimensions.  In  a  properly  designed 
single-phase  motor,  however,  the  relatively  poor  per- 
formance is  not  so  marked.  There  are  certain  consider- 
ations in  the  design  of  the  single-phase  motor  which 
materially  reduce  various  dimensions;  consequently  a 
good  single-phase  motor  is  little  larger  than  a  polyphase 
motor  of  equal  output,  and  its  efficiency  is  not  much 
lower.  These  matters  of  design,  however,  are  beyond 
the  scope  of  this  publication,  and  I  reserve  the  discussion 
of  them  for  some  future  time. 

The  main  object  in  the  preceding  pages  has  been  to 
reduce  to  practice  my  diagram,  and  to  present  it  in  a 
clear  and  easily  understood  form.  If  I  have  succeeded 
in  making  clearer  a  complicated  part  of  our  alternating 
current  theory,  the  purpose  of  this  work  is  fulfilled. 


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